A small issue but one I face every day.
I have this belief that if the temperature of the stove should be no higher than boiling for maximum efficiency. I know that a higher temperature will heat the water to boiling quicker and thus use energy over a shorter period of time, but I can’t help but feel that a lot more energy is wasted in the air, kettle and element during this.
Am I right?
Hmmmm, I don’t think the temp matters.
I think it matters more on the position of the heating coil (I’m assuming it’s electric as a gas flame is going to burn at the same temp no matter what) and the makeup of the pot than it does on the actual temperature.
Think of the water in the pot as an energy sink. Efficiency is measured by how much energy is actually applied to that sink. How much energy does it take to heat the pot? Is it cast iron or aluminum? It’s going to take more energy to heat up the cast iron - that would be a low efficiency; aluminum would be higher. Is the pot full of water? If not, the energy transfered to the pot may then be transfered to the air in the pot rather than the water, again lowering efficiency. Lastly what is the size of the pot, is it smaller than the heating coil? If so you are decreaing efficiency again.
Efficieny is not going to be dependant on temp, but rather the transfer of the energy to the water. So for maximum efficiency:
1.)Use a pot with a high conductivity and low density
2.)Fill it to the brim with water to take maximum use of the pot’s energy
3.)The heating coil should be flat and not bent to maximize the surface area touching the pot
4.)The bottom of the pot should not be smaller than the heating coil, or the coil will lose energy to the air surrounding it (again maximizing suface area touching the pot).
Hon, just put a lid on it.
Assuming the kettle covers the burner completely, I see no reason why this should be show. As a matter of fact, a burner at exactly 100 C would take forever to boil. It takes 1 calorie to heat on gram of water 1 degree and 540 to vaporize one gram once it reaches 100. To actually boil water, a signficant quantity must be vaporizing. In fact, the burners on my stove get red hot and I imagine that doesn’t happen until it is several hundred degrees.
On the other hand, if part of the burner is not covered, then that heat is totally wasted and that has to be a much more significant source of loss than any loss caused by the burner being hotter.
MrsMagoo, we don’t need that around here. We need science! We need experiments and data… We need laboratories and GRA’s! We need government grants! This is a question that must be answered with all due precision.
[size]grumblegrumble freakin’ common sense grumblegrumble[/size]
These are probably more significant but they are also essentially constants in my household.
Some things that might answer my question a little more conclusively for me. Is the heating of the coil and pot linear? IOW is the energy required to make the burner 200C directly proportional to the energy to make it 100C? If a burner is red hot then some of that energy is being converted into light right?
I would imagine there is a ‘most efficient’ energy setting for your stove. Just as cars have a most efficient speed they run at (around 35-40 mph I think). If you run your engine harder you will get where you are going faster but your efficency drops.
Where the ‘sweet spot’ might be for an electric burner is I couldn’t say.
Errata - You will find no clear answer on this - and it will baffle you if you went about mathematically finding an optimum, becuase its a non-linear problem.
But my wag is that if you had a coil at near 100 C, you’ll end up losing more heat to the atmosphere than the heat that lands up in the water.
I’m going to second Whack-a-Mole that there is a most efficient setting. If you set it low enough that the temperature only makes it to 211 F, it’ll never boil, so you’ll just keep putting heat into the room. If you turn it up a little, the water will eventually boil, but take a very long time, losing heat all the while.
OK so the “no higher than boiling” part was a little overstated. I wouldn’t have the patience for it anyway.
But I can boil water in a reasonable amount of time without getting the burners red hot. I figure that light alone must use a few watts extra.
Another idea is the convection. Keep in mind that the burner heats the air around it more quickly as well as the water. I bet that a higher temperature will cause the air around it to move more quickly. Since the room has a relatively unlimited supply of cooler air, it requires more energy for the burner to maintain it’s temperature the higher that it is above the room.
Stoves are inefficent. They lose heat down, up (if the burner is not fully covered) and sideways. If you want more efficency in boiling water get an immersion heater where all the generated heat goes into the liquid. Of course putting a lid on the container and adding insulation heeps heat loss from the liquid down and helps efficency.
The most efficient way I think would be to lower the pressure. I know for useing a gas stove (and camping stove for that matter) that a very low gas flame is most efficient unless you are heating it in very clod temps with a wind that will cool your water faster then a small flame can heat it.
Lowering the pressure lowers the boiling point, though, so if the objective is anything like cooking then we’re heading the wrong way with that.
Start with hot water.
Throw out those rusty blackened steel dishes below the heating coils and buy some new ones. They’re supposed to be mirror-bright REFLECTORS, so the radiation from the bottom of the heating elements gets bounced back up again.
Oh, and also pump all the air out of your kitchen. That will eliminate the energy lost via air convection.
…
I bet that a microwave oven is way more efficient for boiling water. If the dish requires stove-top cooking, then why not use a glass pot, heat it to boiling in the microwave oven, then transfer it to the stove? That’s what I do when making sauces and cream soups. If you use the microwave to first bring the milk to almost boiling, stovetop cooking goes much faster and with less chance of burning.
Very true, and technically answers the question. But boiling water for food cooking would be quite a useless task on say, Everest. And based on the rules of chemistry you can not heat boiling water past its boiling point.
The temperature of a phase changing mass only changes when it is in the process of changing phase. When it is in the solid-liquid stage or in the liquid-gas stage the temp of the matter is changing.
However, when it is in the full liquid phase and has not started to turn gascious (boiling), it will not go above its boiling point at that specific elevation.
So you COULD boil water on Everest (mid 70 degrees C, IIRC), but the heat wouldn’t be great enough to cook the food. It would be just as if you put potatoes in a bucket of water with bubbles comming from little holes in a pipe that goes through a hole in the bottom. You would only succeed in getting them wet and moved around a little bit (possible cleaning invention?).
The Phase change temperature should look rougly something like this IIRC:
| /
| / Gas
(up)| /
l / Liquid/Gas
T | /
E | / Liquid
M | ______________ /
P | / Solid/Liquid
| /
| / Solid
|/_________________________
0 Time------>
The solid liquid section is better know as freezing/melting point and the liquid/gas section is better known as boiling/condensing (rain, condensation etc…).
By raising the pressure the entire section of lines also goes up (this is the principal that allows pressure cookers to work, you get water going signifigantly > 212 degrees C and the pressure only increases that window before it boils, allowing the food to be cooked at a higher temperature before the temperature levels off at the liquid/gas [boiling] section.
And by lowering the pressure the substance melts at a lower temperature, and boils at a lower temperature, making it harder to cook things at Everest heights (or in the vacuum outer space, fwiw).
(Hey… I actually learned something in school…)
Ugh… and my crude attempt of making a diagram has failed.
In a search at google I found this, essentially the same thing that I (badly) attempted to do.
Any heat saved by lowering pressure would be very trivial and, as has been said, would make for poorer cooking. Also, not everybody can go to cook their meals on MT Everest daily so I considered that angle an invalid answer to the OP.
Let’s get to the real question asked in the OP. Contrary to what most people have said, I believe the faster you heat the water, the more energy you save. In fact, if you could could heat it instantly with a sudden jolt of energy, that would be the most efficient. Of course, this is impossible because it would require temperatures in excess of melting points of metals.
A fairly simple mathematical model can be designed which would reflect the behavior of such a system and I am quite sure the result would be what I said.
Any heat saved by lowering the pressure and threfore the boiling temperature would be trivial and, if you take into account the energy required to lower the pressure, then it would be zero or close.
In an immersion heater, my answer (as fast as possible) is, intuitively, clearly and obviously the correct one. Now add slowly some direct losses from heater to outside air and you will see the answer does not change.
All this is assumming an electric heating element where the thermal conductivity in each direction can be considered fixed. In a gas stove, the more the flame extends outside the pot, the more heat lost in the exhaust gases.
They always come to me just after I hit “submit”.
This problem is, conceptually"almost identical to the question we have discussed many times on whether it is better to keep the furnace or AC running or to let the house temperature approximate the outside and then heat/cool when you get home. In both examples, stove and home temperature, time is your enemy. While a stove and a pot seem to have nothing in common with a furnace, the mathematical models of the system are almost identical.
